General Distributivity Theorem

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Theorem

Let $\struct {R, \circ, *}$ be a ringoid.


Then for every pair of sequences $\sequence {a_i}_{1 \mathop \le i \mathop \le m}, \sequence {b_j}_{1 \mathop \le j \mathop \le n}$ of elements of $R$:

$\displaystyle \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$

where:

$\displaystyle \sum_{i \mathop = 1}^m a_i$ is the summation $a_1 \circ a_2 \circ \cdots \circ a_m$
$m$ and $n$ are strictly positive integers: $m, n \in \Z_{> 0}$


Lemmata

The proof requires the following lemmata:

Lemma 1

Let $\left({R, \circ, *}\right)$ be a ringoid.

Then for every sequence $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ of elements of $R$, and for every $b \in R$:

$\displaystyle \left({\sum_{j \mathop = 1}^n a_j}\right) * b = \sum_{j \mathop = 1}^n \left({a_j * b}\right)$


Lemma 2

Let $\struct {R, \circ, *}$ be a ringoid.

Then for every sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$, and for every $b \in R$:

$\displaystyle b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$


Proof

Proof by induction:


For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$


We have that $\struct {R, \circ, *}$ is a ringoid, and so:

$\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$
$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$


Basis for the Induction

$\map P 1$ is the case:

$\displaystyle \forall m \in \Z_{>0}: \paren {\sum_{i \mathop = 1}^m a_i} * b_1 = \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_1}$

This is demonstrated in Lemma 1.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j}$


Then we need to show:

$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}$


Induction Step

This is our induction step:

\(\displaystyle \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j}\) \(=\) \(\displaystyle \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\paren {\sum_{j \mathop = 1}^k b_j} \circ b_{k + 1} }\) Definition of Summation
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} } \circ \paren {\paren {\sum_{i \mathop = 1}^m a_i} * b_{k + 1} }\) as $\struct {R, \circ, *}$ is a ringoid
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} } \circ \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_{k + 1} }\) Basis for the Induction
\(\displaystyle \) \(=\) \(\displaystyle \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j} \circ \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_{k + 1} }\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}\) Associativity of $\circ$ in $\struct {R, \circ, *}$


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\displaystyle \forall m, n \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$

$\blacksquare$


The same result can be obtained by fixing $n$ and using induction on $m$, which requires Lemma 2 to be used for its base case.


Examples

Sum of $j$ from $m$ to $n$ by Sum of $k$ from $r$ to $s$

$\displaystyle \sum_{j \mathop = m}^n \sum_{k \mathop = r}^s j k = \frac 1 4 \left({n \left({n + 1}\right) - \left({m - 1}\right) m}\right) \left({s \left({s + 1}\right) - \left({r - 1}\right) r}\right)$

for $m \le n, r \le s$.


Warning

When using the General Distributivity Theorem, be careful not to conflate the indices.

The following is not a valid argument:

\(\displaystyle \left({\sum_{i \mathop = 1}^n a_i}\right) \left({\sum_{j \mathop = 1}^n \frac 1 {a_j} }\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \frac {a_i} {a_j}\) General Distributivity Theorem
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \sum_{i \mathop = 1}^n \frac {a_i} {a_i}\) Change of Index Variable of Summation
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n 1\) simplification
\(\displaystyle \) \(=\) \(\displaystyle n\)


Also see


Sources